**Question:**

Write each of the sets given below in set7builder from:

(i)

$\mathrm{A}=\left\{1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \frac{1}{25}, \frac{1}{36}, \frac{1}{49}\right\}$

(ii)

$\mathrm{B}=\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\} .$

(iii) $C=\{53,59,61,67,71,73,79\}$.

(iv) $D=\{-1,1\}$.

(v) $E=\{14,21,28,35,42, \ldots ., 98\}$.

**Solution:**

$A=\left\{1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \frac{1}{25}, \frac{1}{36}, \frac{1}{49}\right\}$

$=\left\{\left(\frac{1}{1}\right)^{2},\left(\frac{1}{2}\right)^{2},\left(\frac{1}{3}\right)^{2},\left(\frac{1}{4}\right)^{2},\left(\frac{1}{5}\right)^{2},\left(\frac{1}{6}\right)^{2},\left(\frac{1}{7}\right)^{2}\right\}$

Hence, we may write the set as

$A=\left\{x: x=\frac{1}{n^{2}}, n \in N\right.$ and $\left.1 \leq n \leq 7\right\}$

(ii) $B=\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}$

$=\left\{\frac{1}{1+1}, \frac{2}{4+1}, \frac{3}{9+1}, \frac{4}{16+1}, \frac{5}{25+1}, \frac{6}{36+1}, \frac{7}{49+1}\right\}$

$=\left\{\frac{1}{1^{2}+1}, \frac{2}{2^{2}+1}, \frac{3}{3^{2}+1}, \frac{4}{4^{2}+1}, \frac{5}{5^{2}+1}, \frac{6}{6^{2}+1}, \frac{7}{7^{2}+1}\right\}$

Hence, we may write the set as

$B=\left\{x: x=\frac{n}{n^{2}+1}, n \in N\right.$ and $\left.1 \leq n \leq 7\right\}$

(iii) C = {53, 59, 61, 67, 71, 73, 79}

We know that prime numbers are those numbers which are divisible by 1 and the

number itself.

e.g. $\frac{3}{1}=3$ and $\frac{3}{3}=1$

Here, all the given numbers are consecutive prime numbers greater than 50.

So, C = {x: x is a prime number and 50 < x < 80}

(iv) Here, in set D there are two elements -1 and 1

-1 and 1 are integers

So, the given set can be write as

$D=\{x: x$ is an integer and $-2

(v) $14=7 \times 2$

$21=7 \times 3$

$28=7 \times 4$

$35=7 \times 5$

$42=7 \times 6$

98 = 7 × 14

So, the given set can be write as

E = {x: x = 7n, n ∈ N and 1 ≤ n ≤ 14}